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I am solving this question.

Finding the remainder of

$$\frac{289\times 144^{25}}{71^{71}}$$

This is how I have tried solving it. First it can be simplified to $\frac{17^2 \times 2 ^{50} \times 3^{25}}{71^{71}}$. Now if we use Euler Totient rule we get for $\phi(71^{71}) = 71(1-\frac{1}{71}) = 70 $.

For, $$ \frac{a^{\phi(n)}}{n}$$

I would get remainder as $1$.

I can't proceed from this point. Can someone help me to find out the remainder and where I went wrong?

EDIT : My totient rule was wrong. It's $\phi(71^{71}) = 71^{71}(1-\frac{1}{71}) = 71^{70} \times 70 $.

EDIT #2 : tried solving this in a different way.

$$ \frac{289}{71} \times \frac{144^{25}}{71^{70}}$$

For the first separated division we get $289 \equiv 5 \bmod 71$. Now further simplifying the second, $$\frac{12^{50}}{71^{70}} = \left( \frac{12^5}{71^7} \right)^{10} = \left( \frac{12}{71} \right)^{70} \times \frac{1}{12^{20}}$$

Now I get $12 \equiv -59 \bmod 71$. I'm stuck again here.

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    $\begingroup$ Your Totient formula use is off. For a prime power $p^k$ it should be $p^k(1-1/p)=p^k-p^{k-1}.$ However how does that help in doing the problem? $\endgroup$ – coffeemath Dec 2 '16 at 13:42
  • $\begingroup$ $$\phi(71^{71})= = 71^{71}(1-\frac{1}{71}) = 70 \cdot 71^{70}$$Use Hansel Lemma $\endgroup$ – N. S. Dec 2 '16 at 13:42
  • $\begingroup$ The numerator is smaller than the denominator, so you (just) have to compute the numerator and you can call it a day (or a month) $\endgroup$ – Ross Millikan Dec 2 '16 at 14:58
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${289 \times 144^{25}} \approx 2.6 \times 10^{56} < 2.8 \times 10^{131} \approx 71^{71}$. So the remainder is ${289 \times 144^{25}}$.

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Hint

$17^2 12^{25} 12^{25} < 71^2 71^{25} 71^{25}$

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    $\begingroup$ Should that be $17^2 12^{25} 12^{25}$ on the left-hand side? $\endgroup$ – hmakholm left over Monica Dec 2 '16 at 13:49
  • $\begingroup$ Yes, it should. $\endgroup$ – Jaroslaw Matlak Dec 2 '16 at 14:09
  • $\begingroup$ I didn't understand the hint properly. Also I tried solving this way, $ 289 ≡ 5 $ mod $71$ $ and $ 144 ≡ -2 $ mod $ 71$ $\endgroup$ – Ankit Panda Dec 2 '16 at 14:12
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    $\begingroup$ @sudoankit: $x \bmod y = x $ whenever $0\le x <y$. $\endgroup$ – hmakholm left over Monica Dec 2 '16 at 14:15

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